Ask question

Ask question

All categories

3 years ago

6

An art class is making a mural for their school which has a triangle drawn in the middle. The length of the bottom of the triang

le is x. Another side is 9 more than three times the length of the bottom of the triangle. The last side is 9 more than the bottom of the triangle. Write and simplify an expression for the perimeter of the triangle. What is the expression for the perimeter of the triangle?

1 answer:

jolli1 [7]3 years ago

8 0

Answer:

The perimeter of the triangle can be expressed as 5x + 18.

Step-by-step explanation:

Perimeter of an object is the sum of the length of its sides.

For the required triangle, we have;

length of the base = x

length of the second side = 3x + 9

length of the third side = x + 9

Perimeter of the triangle = length of the base + length of the second side + length of the third side

= x + (3x + 9) + (x + 9)

= x + 3x + 9 + x + 9

= 5x + 18

The perimeter of the triangle can be expressed as 5x + 18.

You might be interested in

Billy earned $150 from a summer job to buy new clothes for school. He found several items he would like to buy, but is trying to

lisabon 2012 [21]

Answer:

Billy went from rich to broke.

Step-by-step explanation:

8 0

3 years ago

2(x - y) in its distributed form is

pochemuha

Answer:

2x-2y

Step-by-step explanation:

2 times x = 2x

2 times -y = -2y

7 0

3 years ago

Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec

julia-pushkina [17]

Answer:

$(2,6,6) \not \in \text{Span}(u,v)$

$(-9,-2,5)\in \text{Span}(u,v)$

Step-by-step explanation:

Let $b=(b_1,b_2,b_3) \in \mathbb{R}^3$ . We have that $b\in \text{Span}\{u,v\}$ if and only if we can find scalars $\alpha,\beta \in \mathbb{R}$ such that $\alpha u + \beta v = b$ . This can be translated to the following equations:

1. $-\alpha + 3 \beta = b_1$

2. $2\alpha+4 \beta = b_2$

3. $3 \alpha +2 \beta = b_3$

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for $\alpha,\beta$ and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = 2$

$2\alpha+4 \beta = 6$

whose unique solutions are $\alpha =1 = \beta$ , but note that for this values, the third equation doesn't hold (3+2 = 5 $\neq$ 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

$-\alpha + 3 \beta = -9$

$2\alpha+4 \beta = -2$

whose unique solutions are $\alpha=3, \beta=-2$ . Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0

3 years ago

Is hunter license a form of a tax

puteri [66]

A hunting license is not a from of tax it is a form of license that allows you to hunt game and only costs each time for renewal of the right
<span />

8 0

3 years ago

Solve for x, 12y+6=6(y+1)

Vanyuwa [196]

For this case we must solve the following equation:

$12y + 6 = 6 (y + 1)$

We apply distributive property on the right side of the equation:

$12y + 6 = 6y + 6$

We subtract 6y on both sides of the equation:

$12y-6y + 6 = 6\\6y + 6 = 6$

We subtract 6 from both sides of the equation:

$6y = 6-6\\6y = 0$

Dividing by 6 on both sides of the equation:

$y = 0$

So, the result is $y = 0$

Answer:

$y = 0$

4 0

3 years ago